Systems
of Non-Linear Equations: Using the Quadratic
Formula (page
6 of 6)

The example below demonstrates
how the Quadratic
Formula is sometimes
used to help in solving, and shows how involved your computations might
get.

Solve the system:

x2
– xy + y2 = 21x2 +
2xy – 8y2 = 0

This system represents
an ellipse and a set of straight lines. If you solve each equation above
for y,
you can enter the "plus-minus" equations into your graphing
calculator to verify this:

x2
– xy + y2 = 21

y2
– xy + (x2 – 21) = 0

x2
+ 2xy – 8y2 = 0

0 = 8y2
– 2xy – x2

As you can see, I used
the Quadratic
Formula in each
case to solve for y
in terms of x.
This gave me equations that I could graph. This technique doesn't come
up that often, but it can be a life-saver when you can't seem to solve
things any other way.

Where did the absolute-value
bars come from? Recall that, technically,
the square root of x2
is the absolute value of x.
That's how I did that simplification in the next-to-last line above.

The absolute value of
x
in the second equation above gives two cases for the values of y:

If x
< 0, then
| x | = –x,
so y
= (x ± 3x)/8 =x/2,
–x/4

If x
> 0, then |
x | = x,
so y
= (x ± 3x)/8 = –x/4,
x/2.

In either case,
y
=–x/4
or y
= x/2.

Since I derived these
"y="
solution-equations
from the second of the original equations, I will plug them into the
first equation to solve for some actual numerical values:

Warning: Do not try to
write the solution points as ""
or "",
because this is not correct. Not all combinations of these x-value
and y-values
are solution points. Don't be sloppy; write the solution out correctly.

By the way, the graph of
the system looks like this:

(To graph the ellipse using
the traditional methods, you would have to do a "rotation of axes",
a process you probably won't see until calculus, if at all.)

There is another way of
proceeding in the above exercise, because the second equation happens
to be factorable. (This factorability is NOT generally true, but
you should try to remember to check, just in case.) If you factor the
second equation and solve for x in
terms of y,
you get:

This last example (the
first way I worked it) is about as complicated as these things ever get.
But if you plug away and work neatly and completely, you should be able
to arrive at the solution successfully. And if you have a graphing calculator
(and the time), try doing a quick graph to verify your answers visually.